Two resistors of resistances $R_{1} = 100 \pm 3 \ \Omega$ and $R_{2} = 200 \pm 4 \ \Omega$ are connected $(a)$ in series,$(b)$ in parallel. Find the equivalent resistance of the $(a)$ series combination,$(b)$ parallel combination. Use for $(a)$ the relation $R = R_{1} + R_{2}$ and for $(b)$ $\frac{1}{R^{\prime}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ and $\frac{\Delta R^{\prime}}{R^{\prime 2}} = \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}}$.

(A) For the series combination,the equivalent resistance is $R = R_{1} + R_{2}$.
$R = (100 + 200) \pm (3 + 4) \ \Omega = 300 \pm 7 \ \Omega$.
$(b)$ For the parallel combination,the equivalent resistance $R^{\prime}$ is given by $\frac{1}{R^{\prime}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$.
$R^{\prime} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{100 \times 200}{100 + 200} = \frac{20000}{300} \approx 66.7 \ \Omega$.
To find the error $\Delta R^{\prime}$,we use $\frac{\Delta R^{\prime}}{R^{\prime 2}} = \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}}$.
$\Delta R^{\prime} = R^{\prime 2} \left( \frac{\Delta R_{1}}{R_{1}^{2}} + \frac{\Delta R_{2}}{R_{2}^{2}} \right) = (66.7)^{2} \left( \frac{3}{100^{2}} + \frac{4}{200^{2}} \right)$.
$\Delta R^{\prime} = 4448.89 \left( \frac{3}{10000} + \frac{4}{40000} \right) = 4448.89 \left( 0.0003 + 0.0001 \right) = 4448.89 \times 0.0004 \approx 1.8 \ \Omega$.
Thus,the equivalent resistance is $66.7 \pm 1.8 \ \Omega$.